Application of Integrals: Application of Integrals (Area Under Curves)

Substitute $y = 2x$ into $y^2 = 4x$: $$(2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies 4x(x - 1) = 0$$ Thus $x = 0$ or $x = 1$. Corresponding $y$-values: when $x = 0, y = 0$; when $x = 1, y = 2$. The intersection points are $(0, 0)$ and $(1, 2)$.

For $x \in [0, 1]$, the curve $y_1 = 2\sqrt{x}$ lies above the line $y_2 = 2x$. $$\text{Area } A = \int_{0}^{1} (y_1 - y_2) \, dx = \int_{0}^{1} (2\sqrt{x} - 2x) \, dx$$

$$A = \left[ 2 \cdot \frac{x^{3/2}}{3/2} - 2 \cdot \frac{x^2}{2} \right]_{0}^{1} = \left[ \frac{4}{3}x^{3/2} - x^2 \right]_{0}^{1}$$ $$= \left( \frac{4}{3}(1) - 1 \right) - (0) = \frac{4}{3} - 1 = \frac{1}{3} \text{ sq. units}$$